Java：将 int 转换为 InetAddress
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Java: convert int to InetAddress
提问by kdt
I have an intwhich contains an IP address in network byte order, which I would like to convert to an InetAddressobject. I see that there is an InetAddressconstructor that takes a byte[], is it necessary to convert the intto a byte[]first, or is there another way?
我有一个int包含网络字节顺序的 IP 地址，我想将其转换为InetAddress对象。我看到有一个InetAddress构造函数需要 a byte[]，是否有必要先将the 转换int为 a byte[]，还是有其他方法？
采纳答案by skaffman
This should work:
这应该有效：
int ipAddress = ....
byte[] bytes = BigInteger.valueOf(ipAddress).toByteArray();
InetAddress address = InetAddress.getByAddress(bytes);
You might have to swap the order of the byte array, I can't figure out if the array will be generated in the correct order.
您可能需要交换字节数组的顺序，我不知道数组是否会以正确的顺序生成。
回答by valli
This may work try
这可能会奏效
public static String intToIp(int i) {
return ((i >> 24 ) & 0xFF) + "." +
((i >> 16 ) & 0xFF) + "." +
((i >> 8 ) & 0xFF) + "." +
( i & 0xFF);
}
回答by Dennis Laumen
Not enough reputation to comment on skaffman's answer so I'll add this as a separate answer.
没有足够的声誉来评论 skaffman 的答案，因此我将其添加为单独的答案。
The solution skaffman proposes is correct with one exception. BigInteger.toByteArray() returns a byte array which could have a leading sign bit.
skaffman 提出的解决方案是正确的，只有一个例外。BigInteger.toByteArray() 返回一个可能有前导符号位的字节数组。
byte[] bytes = bigInteger.toByteArray();
byte[] inetAddressBytes;
// Should be 4 (IPv4) or 16 (IPv6) bytes long
if (bytes.length == 5 || bytes.length == 17) {
// Remove byte with most significant bit.
inetAddressBytes = ArrayUtils.remove(bytes, 0);
} else {
inetAddressBytes = bytes;
}
InetAddress address = InetAddress.getByAddress(inetAddressBytes);
PS above code uses ArrayUtils from Apache Commons Lang.
PS 上面的代码使用来自 Apache Commons Lang 的 ArrayUtils。
回答by aalmeida
I think that this code is simpler:
我认为这段代码更简单：
static public byte[] toIPByteArray(int addr){
return new byte[]{(byte)addr,(byte)(addr>>>8),(byte)(addr>>>16),(byte)(addr>>>24)};
}
static public InetAddress toInetAddress(int addr){
try {
return InetAddress.getByAddress(toIPByteArray(addr));
} catch (UnknownHostException e) {
//should never happen
return null;
}
}
回答by hbr
Tested and working:
测试和工作：
int ip = ... ;
String ipStr =
String.format("%d.%d.%d.%d",
(ip & 0xff),
(ip >> 8 & 0xff),
(ip >> 16 & 0xff),
(ip >> 24 & 0xff));
回答by SHUVA JYOTI KAR
Using Google Guava:
使用谷歌番石榴：
byte[] bytes =Ints.toByteArray(ipAddress);
byte[] bytes =Ints.toByteArray(ipAddress);
InetAddress address = InetAddress.getByAddress(bytes);
InetAddress 地址 = InetAddress.getByAddress(bytes);
回答by user2660727
public InetAddress intToInetAddress(Integer value) throws UnknownHostException
{
ByteBuffer buffer = ByteBuffer.allocate(32);
buffer.putInt(value);
buffer.position(0);
byte[] bytes = new byte[4];
buffer.get(bytes);
return InetAddress.getByAddress(bytes);
}
回答by aij
If you're using Google's Guava libraries, InetAddresses.fromIntegerdoes exactly what you want. Api docs are here
如果您使用的是 Google 的 Guava 库，InetAddresses.fromInteger那么完全可以满足您的需求。Api 文档在这里
If you'd rather write your own conversion function, you can do something like what @aalmeida suggests, except be sure to put the bytes in the right order (most significant byte first).
如果您更愿意编写自己的转换函数，则可以执行类似于@aalmeida 建议的操作，但一定要按正确的顺序放置字节（最重要的字节在前）。
回答by bowman han
public static byte[] int32toBytes(int hex) {
byte[] b = new byte[4];
b[0] = (byte) ((hex & 0xFF000000) >> 24);
b[1] = (byte) ((hex & 0x00FF0000) >> 16);
b[2] = (byte) ((hex & 0x0000FF00) >> 8);
b[3] = (byte) (hex & 0x000000FF);
return b;
}
you can use this function to turn int to bytes;
您可以使用此函数将 int 转换为字节；
